Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $z = \dfrac{-5x - 25}{x^2 + 8x - 20} \times \dfrac{2x + 20}{3x + 15} $
Explanation: First factor the quadratic. $z = \dfrac{-5x - 25}{(x + 10)(x - 2)} \times \dfrac{2x + 20}{3x + 15} $ Then factor out any other terms. $z = \dfrac{-5(x + 5)}{(x + 10)(x - 2)} \times \dfrac{2(x + 10)}{3(x + 5)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac{ -5(x + 5) \times 2(x + 10) } { (x + 10)(x - 2) \times 3(x + 5) } $ $z = \dfrac{ -10(x + 5)(x + 10)}{ 3(x + 10)(x - 2)(x + 5)} $ Notice that $(x + 5)$ and $(x + 10)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac{ -10(x + 5)\cancel{(x + 10)}}{ 3\cancel{(x + 10)}(x - 2)(x + 5)} $ We are dividing by $x + 10$ , so $x + 10 \neq 0$ Therefore, $x \neq -10$ $z = \dfrac{ -10\cancel{(x + 5)}\cancel{(x + 10)}}{ 3\cancel{(x + 10)}(x - 2)\cancel{(x + 5)}} $ We are dividing by $x + 5$ , so $x + 5 \neq 0$ Therefore, $x \neq -5$ $z = \dfrac{-10}{3(x - 2)} ; \space x \neq -10 ; \space x \neq -5 $